Toluene can be partially oxidized to benzoic acid and water in a low pressure vapor phase reaction. Initially, 10 moles of tolue
ne are mixed with 1 mole of air (0.79 moles N2 and 0.21 moles O2). If the reaction is allowed to proceed until all the oxygen has been converted, calculate the mole fractions of each of the components present at the end of the reaction. Ignore the contributions from all possible side reactions.
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<u>Answer:</u> The mass of cryolite produced is 51.48 kg
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For aluminium oxide:</u>
Given mass of aluminium oxide = 12.5 kg = 12500 g (Conversion factor: 1 kg = 1000 g)
Molar mass of aluminium oxide = 101.96 g/mol
Putting values in equation 1, we get:
- <u>For NaOH:</u>
Given mass of NaOH = 55.4 kg = 55400 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
- <u>For HF:</u>
Given mass of HF = 55.4 kg = 55400 g
Molar mass of HF = 20 g/mol
Putting values in equation 1, we get:
For the given chemical reaction:
By Stoichiometry of the reaction:
1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.
So, 122.6 moles of aluminium oxide will react with of sodium hydroxide and
of HF
As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.
Thus, aluminium oxide is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of aluminium oxide produces 2 moles of cryolite
So, 122.6 moles of aluminium oxide will produce = of cryolite
Now, calculating the mass of cryolite by using equation 1:
Molar mass of cryolite = 209.94 g/mol
Moles of cryolite = 245.2 mol
Putting values in equation 1, we get:
Converting this into kilograms, we use the conversion factor:
1 kg = 1000 g
So,
Hence, the mass of cryolite produced is 51.48 kg
